Prove the identity Sin x -cos x /sec x - csc x =sin2x/2?

(sin x -cos x) /(sec x - csc x)
= (sin x - cos x) /( 1/cos x - 1/ sin x) ------(1)

Consider (1/cos x - 1/sin x)
multiply and divide by sin x cos x
= (sin x - cos x) / sin x cos x

Equation (1) becomes:
(sin x - cos x) /[(sin x - cos x)] /sin x cos x]
= sin x cos x
= 2 sin x cos x / 2
= sin 2x / 2

Note;
Use parenthesis

L.H.S.
= (sinx - cosx) / (secx - cscx)
= (sinx - cosx) / [(1 / cosx) - (1 / sinx)]
= (sinx - cosx) / [(sinx / sinx cosx) - (cosx / sinx cosx)]
= (sinx - cosx) / [(sinx - cosx) / sinx cosx]
= (sinx - cosx) × [sinx cosx / (sinx - cosx)]
= sinx cosx
= (2 sinx cosx)/2
= sin2x / 2
= R.H.S.

Hence, (sinx - cosx) / (secx - cscx) = sin2x / 2

Presentation is unclear due to lack of brackets.